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The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is          
  • a)
    0.5 mmol              
  • b)
    1.0 mmol              
  • c)
    2.0 mmol              
  • d)
    2.5 mmol
Correct answer is option 'B'. Can you explain this answer?
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The standard reduction potential of a silver chloride electrode is 0.2...
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The standard reduction potential of a silver chloride electrode is 0.2...
To determine the maximum amount of AgCl that can dissolve in the given solution, we need to consider the reduction potentials of the silver chloride and silver electrodes.

1. Half-Reactions:
The reduction half-reaction for the silver chloride electrode can be represented as follows:
AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq)

The reduction half-reaction for the silver electrode can be represented as follows:
Ag⁺(aq) + e⁻ → Ag(s)

2. Standard Reduction Potentials:
The standard reduction potential for the silver chloride electrode is given as 0.2 V, and the standard reduction potential for the silver electrode is given as 0.79 V.

3. Cell Potential:
The cell potential, E°cell, for the cell can be determined by subtracting the reduction potential of the anode from the reduction potential of the cathode.

E°cell = E°cathode - E°anode

In this case, the silver electrode is the cathode and the silver chloride electrode is the anode.

E°cell = 0.79 V - 0.2 V
E°cell = 0.59 V

4. Nernst Equation:
The Nernst equation relates the standard reduction potential to the concentration of ions in solution. It can be used to calculate the cell potential under non-standard conditions.

Ecell = E°cell - (0.0592/n)log(Q)

Where:
Ecell = cell potential under non-standard conditions
E°cell = standard cell potential
n = number of moles of electrons transferred in the balanced equation
Q = reaction quotient = [products]/[reactants]

In this case, since the concentration of AgCl is being asked, we consider the reaction quotient as the concentration of Ag⁺ and Cl⁻ ions.

5. Calculation:
Given that the volume of the solution is 106 L and the concentration of AgNO3 is 0.1 M, we can calculate the maximum amount of AgCl that can dissolve.

Q = [Ag⁺][Cl⁻]
Q = (0.1 M)(0.1 M)
Q = 0.01

Substituting the values into the Nernst equation:

0.59 V = 0.59 V - (0.0592/1)log(0.01)

Simplifying:

0 = - (0.0592)log(0.01)

Since the logarithm of 0.01 is -2, we have:

0 = - (0.0592)(-2)
0 = 0.1184

This is not possible, as the equation cannot equal zero.

Therefore, the maximum amount of AgCl that can dissolve in the given solution is 0 mmol or option 'B'.
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The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution isa)0.5 mmolb)1.0 mmolc)2.0 mmold)2.5 mmolCorrect answer is option 'B'. Can you explain this answer?
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